Question

the sum of three consecutive numbers is 41 less than five times the largest number. find the largest number

Answer 1

The largest number is:

19

Explanation:

Let the three consecutive numbers be:

x , x+1 and x+2 respectively

Now, it is given that:

The sum of three consecutive numbers is 41 less than five times the largest number.

i.e.

`x+(x+1)+(x+2)=5(x+2)-41`

i.e.

`x+x+x+1+2=5x+10-41\\\\3x+3=5x-31\\\\5x-3x=3+31\\\\2x=34\\\\x=17`

Hence, the smallest number is: 17

The middle number is: 17+1=18

and the largest number is: 17+2=19

Answer 2

Hi,

You can write your problem as an equation.

x - 2 + x - 1 + x = 5x - 41 ,

x being the largest number.
So,

3x - 3 = 5x - 41
3x - 5x = -41 +3
-2x = -38
2x = 38
x = 38/2 = 19

The largest number is 19.

Hope this helps! If my answer was not clear enough or you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistakes.